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Nina [5.8K]
4 years ago
7

One of the pieces of evidence supporting energy quantization was the line spectra of elements. Why does this demonstrate energy

quantization?
a) Photons are packets of energy that act both as a particle and a wave.
b) The spectra is continuous and shows emission at all wavelengths.
c) There are sharp emission lines demonstrating discrete energy levels.
d) All compounds emit light in the ultraviolet and visible regions specifically.​
Chemistry
1 answer:
bezimeni [28]4 years ago
8 0

Answer:

c) There are sharp emission lines demonstrating discrete energy levels.

Explanation:

When an element emits energy in the form of radiation, it produces a spectrum of colors on a photographic plate. This spectrum can either be continuous or discrete. In continuous spectrum the spectrum continues without any discrimination between two regions. This represents the continuous emission of radiation, and thus the continuous emission of energy without any break.

On the other hand, the line spectrum consists of discrete and sharp lines, which shows the emission of radiation in a certain amount in a certain time, with a break between emission. Hence, the line spectra supports the quantization of energy.

The correct option is:

<u>c) There are sharp emission lines demonstrating discrete energy levels.</u>

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What quantity of copper is deposited by the same quantity of electricity that deposited 9g of aluminum
klio [65]

Answer:

Mass of copper deposited = 31.75 g

Explanation:

According to Faraday's second law of electrolysis, when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of the elements.

From this law, it can be seen that the higher the charge, the lower the number of moles of a given element deposited.

Number of moles of aluminium in 9 g of aluminium = mass / molar mass

Molar mass of aluminium = 27 g

Number of moles of aluminium = 9/27 = 1/3 moles

Charge on aluminium ion = +3

3 moles of electrons will discharge 1 mole of aluminium,

1 mole of electrons will discharge 1/3 moles of aluminium

Number of moles of electrons involved = 1 mole of electrons

Charge on copper ion = +2

1 mole of electrons will discharge 1/2 moles of copper.

Mass of 1/2 moles of copper = number of moles × molar mass of copper

Molar mass of copper = 63.5 g

Mass of copper deposited = 1/2 × 63.5 = 31.75 g

3 0
3 years ago
6th Grade Science question on Mass, Volume, and Density!! I've tried and tried and I can't figure it out!​
Sonbull [250]
If you want to find density, do mass divided by the volume and there’s your answer!
8 0
3 years ago
The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C
Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 460^oC = 5.8\times 10^{-6}s^{-1}

K_2 = rate constant at T_2 = 4\times K_1

Ea = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 460^oC=273+460=733K

T_2 = final temperature = ?

Now put all the given values in this formula, we get:

\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]

T_2=791.1K

Therefore, the final temperature would be, 791.1 K

4 0
3 years ago
A lab technician needs a 40% solution of a certain chemical for an experiment, but the lab that she works in only has a 20% solu
Snezhnost [94]

To determine the volume of both concentration of the solution to make another concentration of solution, we need to set up two equations since we have two unknowns. <span>

For the first equation, we do a mass balance:

mass of 50% solution + mass of 20% solution = mass of 40% solution
M1 + M2 = M3

For the second equation, we do a component balance,</span>

<span>
M1(50%) + M2(20%) = M3(40%)
.50M1 + .20M2 = .40M3

To determine the ratio, we assume we have to make a 100 g of the 40% solution. So, the equation would change to</span>

<span>
</span>

<span>M1 + M2 = 100</span>

.50M1 + .20M2 = (100)(.40) = 40


Solving for M1 and M2,

M1 = 66.67 g

M2 = 33.33 g


So, the ratio of the 20% and the 50% would be approximately 33.33/66.67 = 0.5.

8 0
3 years ago
Which reaction occurs at the cathode of a galvanic cell that has an aluminum electrode in an electrolyte with aluminum ions and
Elena L [17]

Answer:

{ \bf{Cu {}^{2 + } + 2e \:  → \: Cu }}

6 0
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